Question

What is the tension in a wire supporting a $1250.0N$ underwater camera submerged in water? The volume of the camera is $8.3 \times {10^{ - 2}}{m^3}$?

Answer 1

To solve this question we will first trace all the forces applied on the object from all directions and equate it to zero as the net force on the object is zero.

Complete step by step solution:
Since the camera is supported and not accelerating, the number of the forces on it should be zero. The friction in the string that lifts the camera and the force of buoyancy are two forces working upward (which is why things float in water). The downward force is the weight of the camera due to its mass. The total should be zero.
The equation below is obtained by treating the upward forces as positive and the downward forces as negative (so that they will mathematically cancel).
$\text{F = Tension + Buoyancy - weight}$
To calculate the value of Buoyancy , we use the following formula
$Buoyancy = p.v.g$
Given the amount of volume displaced by the camera = $8.3 \times 10^{-2}m^3$
Where, $p$ is the density of the water, $v$ is the volume (of the amount of water displaced by the camera) and $g$ is the acceleration due to gravity. Substituting these values into the equation we get
[Density of the water is $1000 kg/m^3$]
$\text{Buoyancy} = 1000 \times 0.083 \times 9.8 \\\ \text{Buoyancy} = 813.4N \\\$
Now, We also know that
$F = 0N \\\ \text{Weight} = 1250N \\\$
Substituting these values to the main equation we get
$0 = \text{Tension} + 813.4 - 1250 \\\ 0 = \text{Tension} - 436.6N \\\ \text{Tension} = 436.6N \\\$
Hence, the tension is $436.6N$ in the upward direction.

Note:
Tension is defined in physics as the axial pulling force conveyed by a thread, chain, cable, or other one-dimensional continuous material, or by each end of a rod, truss component, or other three-dimensional object.